Determine x(2) if x(n) = (1/3)^ ^(-n). What is the regions of convergence of 2?
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![Determine x(2) if x(n) = (13)"^ ^(-n).
What is the regions of convergence of 2
2?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe704cb26-71bb-4db7-be41-af84a74c6da3%2F98ec3b2b-3b17-4686-917f-619ea4b39136%2Foywfu1p_processed.jpeg&w=3840&q=75)
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- express n-1 Σ((k^2)/n) k=1 in closed form.I had tried Xn=(-1)^nbut since 2((-1)^n)^n =2(-1)^(n^2) (which is not always 1) it doesn't converge to some real number.For a positive integer n, prove that summation d/n mu^(2)(d) phi(d) = n / phi(n). [Hint: Both sides of the equation are multiplicative functions].
- what is the summation where k = 1 of e^-2kUse this function to show that 1+(1/3^2)+(1/5^2)+...=pi^2/8 f(x)=.5-(Summation from n=0 to infinity[(4cos((2n-1)(pi)(x)))/(pi^2*(2n-1)^2)] or probably more useful it's problem 18 on this pdf https://www.stewartcalculus.com/data/CALCULUS%206E%20Early%20Transcendentals/upfiles/topics/6et_at_01_fs_stu.pdfB please, for at least 3 and 4 T^T
- i got =(4BC^(-1))*(1/3)CSolve for the sum for the following: ∞ Σ [(-1^n)*π^(2n+1)]/[(4^(2n+1))*(2n+1)] n=0To find the unique solution p^∗ ∈ [0, 1] of the equation x^3 + 6x^2 − 4 = 0, rewrite the equation in the fixed-point form x = g(x) with two different choices of g, such that the sequence {pn} from the fixed-point iteration pn = g(pn−1) is expected to converge to p^∗ when p0 is sufficiently close to p^∗(but not equal to p^∗). Explain why your choices of g would work.
- given the function f(x) = 6x^2 + 2x - 1 from [1,9] find the upper sum using four rectangles.3)An electronic components firm launches a new product on 1 January. During the following year arough estimate of the number of orders, O, received t days after the launch is given by O= 2t^2 − 0.004t^3 (a) What is the maximum number of orders received on any one day of the year?(b) After how many days does the firm experience the greatest increase in orders?Does f(x) = 3^n+2 / 5^n converge or diverge? If it converges find the limit.